The figure below shows the radial velocity of a star plotted as a function of time over the course of 20 days. Where is the planet located in its orbit on day 1 (the start of the graph)? 20 Orbit of star 10- Earth 10+ Time -20 Orbit of planet Radial Velocity
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- Two planets, planet A and planet B, have the same surface gravity. However, planet B has twice the radius of planet A. How does the mass of planet B compare to the mass of planet A? The mass of planet B is four times the mass of planet A. The mass of planet B is equal to the mass of planet A. The mass of planet B is one-half the mass of planet A. The mass of planet B is twice the mass of planet A. The mass of planet B is one-fourth the mass of planet A.What is the apparent magnitude of the sun as seen from Saturn about 10 AU away? Appear to magnitude of Saturn =A planet of mass m= 8.45 x 1024 kg is orbiting in a circular path a star of mass M= 6.95 x 1029 kg. The radius of the orbit is R= 3.15 x 107km. What is the orbital pperiod (in Earth days) of the planet Pplanet? Express your answer to three significant figures. Pplanet = ? days
- The Algol binary system consists of a 3.7 Msun star and a 0.8 Msun star with an orbital period of 2.87 days. Using Newton’s version of Kepler’s Third Law, calculate the distance, a, between the two stars. Compare that to the size of Betelgeuse (you’ll need to look that up). Newton’s Version of Kepler’s Law: (M1 + M2) P2 = (4p2 /G) a3 Rearrange the equation to solve for a. Pi, p, is equal to 3.14. IMPORTANT NOTE: Google the value of G (the Universal Gravitational Constant) or look it up in your text. NOTICE THE UNITS. You must convert every distance and time in your equation to the same units, otherwise, you’ll get an incorrect answer. That means you must convert distances to meters, solar masses to kilograms, and time to seconds. When you compare your value to the size of Betelgeuse, it will also help that they are in the same units.Assume that there are 2 stars - Aqua and Ruby. They are in orbit around a massive star which is named as Star Ai that has a mass 5.98 x 1028 kg. The orbit of Ruby is circle and has a period of 1.430 yrs. The orbit of Aqua is an ellipse. At its periapsis, the distance of Star Aqua is twice as far from Star Ai than Star Ruby is. Solve how far is Star Aqua's apoapsis if Star Aqua is considered 3.250 times faster in periapsis than in apoapsis. Note: You can ignore the gravitational interaction that exists between Star Aqua and Star Ruby. Use the figure to understand the problem better. Aqua periapsis Choices: a. 1.9x10¹¹ m b. 3.8x10¹1 m c. 3.6x10¹0 m d. 1.2x10¹¹ m Ruby Ai apoapsisIf we view a star now, and then view it again 6 months later, our position will have changed by the diameter of the Earth's orbit around the sun. For nearby stars (within 100 light-years or so), the change in viewing location is sufficient to make the star appear to be in a slightly different location in the sky. Half of the angle from one location to the next is known as the parallax angle (see figure). Parallax can be used to measure the distance to the star. An approximate relationship is given by d = 3.26 p , where d is the distance in light-years, and p is the parallax measured in seconds of arc. Vega is a star that has a parallax angle of 0.13 second. How far is Vega from the sun? Note: Parallax is used not only to measure stellar distances. Our binocular vision actually provides the brain with a parallax angle that it uses to estimate distances to objects we see. (Round your answer to two decimal places.) light-years
- Q2Use a distance of R = 1.48x10^11 meters for the distance between the earth and the sun. Use a mass of 1.99x10^30 kg to be 1 solar mass. For each of the different sun masses (as values of solar mass, aka 0.5 solar masses = 1x10^30 kg), as outlined in the lecture, calculate the period of the earth's orbit in days using Kepler's law for circular orbits (I double-checked it with these values and it works) and also calculate the corresponding orbital velocity of the earth. Questions: 1.) Using these values, and 6x10^24 kg for the mass of the earth, what is the strength of the gravitational force between the earth and the sun? 2.) If the earth were twice as far from the sun, what would be its period of orbit? 3.) Mars orbits the sun at a distance of 2.18x10^11 meters. How long is a Martian year, using Kepler's law for circular orbits?The table below presents the semi-major axis (a) and Actual orbital period for all of the major planets in the solar system. Cube for each planet the semi-major axis in Astronomical Units. Then take the square root of this number to get the Calculated orbital period of each planet. Fill in the final row of data for each planet. Table of Data for Kepler’s Third Law: Table of Data for Kepler’s Third Law: Planet aau = Semi-Major Axis (AU) Actual Planet Calculated Planet Period (Yr) Period (Yr) __________ ______________________ ___________ ________________ Mercury 0.39 0.24 Venus 0.72 0.62 Earth 1.00 1.00 Mars 1.52 1.88 Jupiter…
- Plant A orbits a star at a distance R from the center of the star. Planet B circles the same star with a period eight times that of Planet A. Planet B orbits the star at a distance of. a. 4R b. 1/8R c. 8R c. 1/4RWhat would be the distance to the nearest star, Alpha Centauri (4.4 light-years away). (Hint: Find the distance to Alpha Centauri in units of AU.) Express your answer to two significant figures and include the appropriate units.A new mystery planet is detected around our Sun. We measure it’s position relative to the Sun to be 2 AU at perihelion and 6 AU at aphelion. What is the semi-major axis of this planet’s orbit (in AU)? With that information, what is the orbital period of that planet(in years)?